Blog · Calculus guides · 3 November 2026 · 4 min read

By the Asymptote team. Every answer on this page is checked by the same machine that grades the game.

Area Between Curves: Formula, 3 Examples & Practice Problems

The area between two curves is the integral of the top curve minus the bottom curve: $A = \int_{a}^{b} (f (x) - g (x)) d x$ , where $f (x) \geq g (x)$ on $[a, b]$ . The limits are usually where the curves meet, found by setting $f (x) = g (x)$ . If the curves cross in between, split the integral at each crossing so every piece is top minus bottom.

The formula

A = \int_{a}^{b} (top (x) - bottom (x)) d x

The area under the top curve, minus the area under the bottom one. Nothing more mysterious than cutting a shape out of a bigger shape.

When to use it

Use this whenever a question asks for the area enclosed by two or more curves, or between a curve and a line. It is the area under a curve with a second curve taking the place of the $x$ -axis, and it is also the first step of every volume problem: disks and washers and shells start from exactly this region. If the curves are easier to write as $x$ in terms of $y$ , integrate right minus left in $d y$ instead.

How to do it, step by step

Find where the curves meet. Set them equal and solve. Those $x$ -values are usually your limits.
Decide which curve is on top. Pick any test point between the limits and compare the two values. Do not trust the way the equations look.
Check for crossings. If the curves meet again between your limits, the top and bottom swap there, and you need a separate integral for each piece.
Integrate top minus bottom and add the pieces. Every piece should come out positive. If one is negative, its top and bottom are the wrong way round.

Worked examples

Example 1: A line and a parabola

Easy

Find the area enclosed by $y = 4 x$ and $y = x^{2}$ .

y = 4 x and y = x^{2}

Set them equal: $x^{2} = 4 x$ , so $x (x - 4) = 0$ , giving $x = 0$ and $x = 4$ .

Test $x = 1$ : the line gives $4$ , the parabola gives $1$ . The line is on top.

A = \int_{0}^{4} (4 x - x^{2}) d x = [2 x^{2} - \frac{x ^{3}}{3}]_{0}^{4} = 32 - \frac{64}{3}

Read it as the triangle under the line, which has area $32$ , minus the part under the parabola, $\frac{64}{3}$ . That is the formula doing exactly what the caption promised.

Answer

A = \frac{32}{3}

Example 2: Limits that need a quadratic

Medium

Find the area enclosed by $y = 2 - x^{2}$ and $y = x$ .

y = 2 - x^{2} and y = x

Set them equal: $2 - x^{2} = x$ , so $x^{2} + x - 2 = 0$ , which factors as $(x + 2) (x - 1) = 0$ . The curves meet at $x = - 2$ and $x = 1$ .

Test $x = 0$ : the parabola gives $2$ , the line gives $0$ . The parabola is on top, so the integrand is $(2 - x^{2}) - x$ .

A = \int_{- 2}^{1} (2 - x^{2} - x) d x = [2 x - \frac{x ^{3}}{3} - \frac{x ^{2}}{2}]_{- 2}^{1}

= (2 - \frac{1}{3} - \frac{1}{2}) - (- 4 + \frac{8}{3} - 2) = \frac{7}{6} + \frac{10}{3}

The bracket for $x = - 2$ is negative, so subtracting it adds. Brackets around the lower limit save marks here.

Answer

A = \frac{9}{2}

Example 3: Curves that cross

Harder

Find the total area enclosed by $y = x^{3}$ and $y = x$ .

y = x^{3} and y = x

Set them equal: $x^{3} = x$ , so $x (x - 1) (x + 1) = 0$ . They meet three times, at $x = - 1$ , $0$ and $1$ , which makes two separate regions.

Test each one. At $x = \frac{1}{2}$ , the line gives $\frac{1}{2}$ and the cubic $\frac{1}{8}$ , so the line is on top on $[0, 1]$ . At $x = - \frac{1}{2}$ it is the other way round: the cubic is on top on $[- 1, 0]$ .

\int_{0}^{1} (x - x^{3}) d x = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}

Both curves are odd functions, so the left region is the right one rotated half a turn about the origin, and has the same area. Add the two pieces.

If you had integrated $x - x^{3}$ straight from $- 1$ to $1$ , the two pieces would cancel and you would get $0$ : a region with no area, which should make anyone suspicious.

Answer

A = 2 \cdot \frac{1}{4} = \frac{1}{2}

Area between curves practice problems

Three to try, easy to harder. Type an answer and it is checked instantly by the same checker that grades the game; any correct form counts. Stuck? The answer and the first step are under each one.

Problem 1

Easy

Find the area enclosed by the two curves:

y = 6 x and y = 3 x^{2}

Stuck? Show the answer

Answer

4

How to start

They meet where $3 x^{2} = 6 x$ , at $x = 0$ and $x = 2$ . The line is on top, so integrate $6 x - 3 x^{2}$ from $0$ to $2$ .

Problem 2

Medium

Find the area enclosed by the two parabolas:

y = 9 - x^{2} and y = x^{2} - 9

Stuck? Show the answer

Answer

72

How to start

They meet at $x = \pm 3$ . Top minus bottom is $(9 - x^{2}) - (x^{2} - 9) = 18 - 2 x^{2}$ .

Problem 3

Harder

Find the total area enclosed by the curve and the $x$ -axis:

y = 4 x - x^{3} and y = 0

Stuck? Show the answer

Answer

8

How to start

The curve crosses the axis at $x = - 2, 0, 2$ , so there are two regions, one above the axis and one below. Work out $\int_{0}^{2} (4 x - x^{3}) d x$ and use the symmetry.

Want more? Asymptote has 380 more area between curves problems, free, with no account.

Common mistakes

Bottom minus top. You get the right size with the wrong sign. An area cannot be negative, so a negative answer is a free warning.
Integrating across a crossing. If the curves swap over between your limits, one integral counts part of the area as negative and cancels it out. Split at every crossing.
Guessing the limits. The limits come from solving top = bottom, not from the question's diagram. Solve the equation, even when the picture looks obvious.
Losing a sign at the lower limit. $F (b) - F (a)$ with a negative $a$ is where most arithmetic slips happen. Bracket $F (a)$ before you subtract it.

Practice it now

No clock, three strikes

Area problems arrive in Diverge around d6-d7: a line and a parabola, two parabolas, a parabola cut by a line, and a cubic against the $x$ -axis that needs the split from the third example. Answers are exact numbers, so roots like $5$ stay as roots.

Play Diverge See it on the formula sheet

FAQ

How do you know which curve is on top?: Pick a number between the limits, plug it into both, and see which is bigger. One test point per region is enough, as long as the curves do not cross inside it.
Can the area between curves be negative?: No. A negative result means top and bottom are swapped somewhere. A definite integral can be negative; an area cannot.
When should I integrate with respect to $y$ ?: When the region is bounded on the left and right by curves of the form $x = g (y)$ , such as $x = y^{2}$ . Then the area is the integral of right minus left, $d y$ .
What if the curves cross more than once?: Find every crossing, split the interval at each one, and integrate top minus bottom on each piece, then add.
Where can I find area between curves practice problems?: Right on this page: the practice problems section has three area between curves problems, easy, medium and harder, with the answer and the first step under each one, and a box that checks what you type with the same checker that grades Asymptote. The game has 380 more area between curves problems after that, free and with no account: try Diverge, which has no clock.

Area Between Curves: Formula, 3 Examples & Practice Problems

When to use it

Use this whenever a question asks for the area enclosed by two or more curves, or between a curve and a line. It is the area under a curve with a second curve taking the place of the

x

-axis, and it is also the first step of every volume problem: disks and washers and shells start from exactly this region. If the curves are easier to write as

x

in terms of

y

, integrate right minus left in

d y

instead.

How to do it, step by step

Find where the curves meet. Set them equal and solve. Those

x

-values are usually your limits.

Decide which curve is on top. Pick any test point between the limits and compare the two values. Do not trust the way the equations look.

Check for crossings. If the curves meet again between your limits, the top and bottom swap there, and you need a separate integral for each piece.

Integrate top minus bottom and add the pieces. Every piece should come out positive. If one is negative, its top and bottom are the wrong way round.

Worked examples

Example 1: A line and a parabola

Easy

Find the area enclosed by $y = 4 x$ and $y = x^{2}$ .

y = 4 x and y = x^{2}

Set them equal: $x^{2} = 4 x$ , so $x (x - 4) = 0$ , giving $x = 0$ and $x = 4$ .

Test $x = 1$ : the line gives $4$ , the parabola gives $1$ . The line is on top.

A = \int_{0}^{4} (4 x - x^{2}) d x = [2 x^{2} - \frac{x ^{3}}{3}]_{0}^{4} = 32 - \frac{64}{3}

Read it as the triangle under the line, which has area $32$ , minus the part under the parabola, $\frac{64}{3}$ . That is the formula doing exactly what the caption promised.

Answer

A = \frac{32}{3}

Example 2: Limits that need a quadratic

Medium

Find the area enclosed by $y = 2 - x^{2}$ and $y = x$ .

y = 2 - x^{2} and y = x

Set them equal: $2 - x^{2} = x$ , so $x^{2} + x - 2 = 0$ , which factors as $(x + 2) (x - 1) = 0$ . The curves meet at $x = - 2$ and $x = 1$ .

Test $x = 0$ : the parabola gives $2$ , the line gives $0$ . The parabola is on top, so the integrand is $(2 - x^{2}) - x$ .

A = \int_{- 2}^{1} (2 - x^{2} - x) d x = [2 x - \frac{x ^{3}}{3} - \frac{x ^{2}}{2}]_{- 2}^{1}

= (2 - \frac{1}{3} - \frac{1}{2}) - (- 4 + \frac{8}{3} - 2) = \frac{7}{6} + \frac{10}{3}

The bracket for $x = - 2$ is negative, so subtracting it adds. Brackets around the lower limit save marks here.

Answer

A = \frac{9}{2}

Example 3: Curves that cross

Harder

Find the total area enclosed by $y = x^{3}$ and $y = x$ .

y = x^{3} and y = x

Set them equal: $x^{3} = x$ , so $x (x - 1) (x + 1) = 0$ . They meet three times, at $x = - 1$ , $0$ and $1$ , which makes two separate regions.

\int_{0}^{1} (x - x^{3}) d x = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}

Both curves are odd functions, so the left region is the right one rotated half a turn about the origin, and has the same area. Add the two pieces.

If you had integrated $x - x^{3}$ straight from $- 1$ to $1$ , the two pieces would cancel and you would get $0$ : a region with no area, which should make anyone suspicious.

Answer

A = 2 \cdot \frac{1}{4} = \frac{1}{2}

Area between curves practice problems

Three to try, easy to harder. Type an answer and it is checked instantly by the same checker that grades the game; any correct form counts. Stuck? The answer and the first step are under each one.

Problem 1

Easy

Find the area enclosed by the two curves:

y = 6 x and y = 3 x^{2}

Stuck? Show the answer

Answer

4

How to start

They meet where $3 x^{2} = 6 x$ , at $x = 0$ and $x = 2$ . The line is on top, so integrate $6 x - 3 x^{2}$ from $0$ to $2$ .

Problem 2

Medium

Find the area enclosed by the two parabolas:

y = 9 - x^{2} and y = x^{2} - 9

Stuck? Show the answer

Answer

72

How to start

They meet at $x = \pm 3$ . Top minus bottom is $(9 - x^{2}) - (x^{2} - 9) = 18 - 2 x^{2}$ .

Problem 3

Harder

Find the total area enclosed by the curve and the $x$ -axis:

y = 4 x - x^{3} and y = 0

Stuck? Show the answer

Answer

8

How to start

The curve crosses the axis at $x = - 2, 0, 2$ , so there are two regions, one above the axis and one below. Work out $\int_{0}^{2} (4 x - x^{3}) d x$ and use the symmetry.

Common mistakes

Bottom minus top. You get the right size with the wrong sign. An area cannot be negative, so a negative answer is a free warning.

Integrating across a crossing. If the curves swap over between your limits, one integral counts part of the area as negative and cancels it out. Split at every crossing.

Guessing the limits. The limits come from solving top = bottom, not from the question's diagram. Solve the equation, even when the picture looks obvious.

Losing a sign at the lower limit.

F (b) - F (a)

with a negative

a

is where most arithmetic slips happen. Bracket

F (a)

before you subtract it.

Practice it now

No clock, three strikes

Play Diverge See it on the formula sheet

FAQ

How do you know which curve is on top?

Pick a number between the limits, plug it into both, and see which is bigger. One test point per region is enough, as long as the curves do not cross inside it.

Can the area between curves be negative?

No. A negative result means top and bottom are swapped somewhere. A definite integral can be negative; an area cannot.

When should I integrate with respect to

y

When the region is bounded on the left and right by curves of the form

x = g (y)

, such as

x = y^{2}

. Then the area is the integral of right minus left,

d y

What if the curves cross more than once?

Find every crossing, split the interval at each one, and integrate top minus bottom on each piece, then add.

Where can I find area between curves practice problems?

Right on this page: the practice problems section has three area between curves problems, easy, medium and harder, with the answer and the first step under each one, and a box that checks what you type with the same checker that grades Asymptote. The game has 380 more area between curves problems after that, free and with no account: try Diverge, which has no clock.

Area Between Curves: Formula, 3 Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A line and a parabola

Example 2: Limits that need a quadratic

Example 3: Curves that cross

Area between curves practice problems

Common mistakes

FAQ

Keep going

Area Between Curves: Formula, 3 Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A line and a parabola

Example 2: Limits that need a quadratic

Example 3: Curves that cross

Area between curves practice problems

Common mistakes

FAQ

Keep going